BZOJ 3685: 普通van Emde Boas树( 线段树 )

建颗权值线段树就行了...连离散化都不用...

没加读入优化就TLE, 加了就A掉了...而且还快了接近1/4....

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#include<bits/stdc++.h>

   

#define rep(i, n) for(int i = 0; i < n; i++)

#define clr(x, c) memset(x, c, sizeof(x))

#define M(l, r) (((l) + (r)) >> 1)

 

using namespace std;

 

const int maxn = 1000009; 

 

struct Node {

Node *l, *r;

bool s;

Node() {

s = false;

}

inline void update() {

s = l->s || r->s;

}

} pool[maxn << 1], *pt = pool, *root;

 

void build(Node* t, int l, int r) {

if(r > l) {

int m = M(l, r);

build(t->l = pt++, l, m);

build(t->r = pt++, m + 1, r);

}

}

 

int n, v, type;

 

void modify(Node* t, int l, int r) {

if(l == r) {

if(type == 1 && !t->s) t->s = true;

if(type == 2 && t->s) t->s = false;

} else {

int m = M(l, r);

v <= m ? modify(t->l, l, m) : modify(t->r, m + 1, r);

t->update();

}

}

 

int query(Node* t, int l, int r) {

if(!t->s) return 0;

if(l == r) return l;

int m = M(l, r);

return type == 3 ? (t->l->s ? query(t->l, l, m) : query(t->r, m + 1, r)) : 

                  (t->r->s ? query(t->r, m + 1, r) : query(t->l, l, m));

}

 

int succ(Node* t, int l, int r) {

if(!t->s) return 0;

if(l == r) return l <= v ? 0 : l;

int m = M(l, r);

if(v >= m) return succ(t->r, m + 1, r);

int ans = succ(t->l, l, m);

return ans ? ans : succ(t->r, m + 1, r);

}

 

int pred(Node* t, int l, int r) {

if(!t->s) return 0;

if(l == r) return l >= v ? 0 : l;

int m = M(l, r);

if(m + 1 >= v) return pred(t->l, l, m);

int ans = pred(t->r, m + 1, r);

return ans ? ans : pred(t->l, l, m);

}

 

int find(Node* t, int l, int r) {

if(!t->s) return -1;

if(l == r) return 1;

int m = M(l, r);

return v <= m ? find(t->l, l, m) : find(t->r, m + 1, r);

}

 

inline void read(int &t) {

t = 0;

char c = getchar();

for(; !isdigit(c); c = getchar());

for(; isdigit(c); c = getchar())

   t = t * 10 + c - '0';

}

 

int main() {

freopen("test.in", "r", stdin);

freopen("test.out", "w", stdout);

int m;

cin >> n >> m;

n++;

build(root = pt++, 1, n);

while(m--) {

read(type);

switch(type) {

case 1 : read(v); v++; modify(root, 1, n); break;

case 2 : read(v); v++; modify(root, 1, n); break;

case 3 : printf("%d\n", query(root, 1, n) - 1); break;

case 4 : printf("%d\n", query(root, 1, n) - 1); break;

case 5 : read(v); v++; printf("%d\n", pred(root, 1, n) - 1); break;

case 6 : read(v); v++; printf("%d\n", succ(root, 1, n) - 1); break;

case 7 : read(v); v++; printf("%d\n", find(root, 1, n)); break;

default : break;

}

}

return 0;

}

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3685: 普通van Emde Boas树

Time Limit: 9 Sec  Memory Limit: 128 MB
Submit: 616  Solved: 216
[Submit][Status][Discuss]

Description

设计数据结构支持:
1 x  若x不存在,插入x
2 x  若x存在,删除x
3    输出当前最小值,若不存在输出-1
4    输出当前最大值,若不存在输出-1
5 x  输出x的前驱,若不存在输出-1
6 x  输出x的后继,若不存在输出-1
7 x  若x存在,输出1,否则输出-1

Input

第一行给出n,m 表示出现数的范围和操作个数
接下来m行给出操作
n<=10^6,m<=2*10^6,0<=x<n

Output

Sample Input

10 11
1 1
1 2
1 3
7 1
7 4
2 1
3
2 3
4
5 3
6 2

Sample Output

1
-1
2
2
2
-1

HINT

Source

By Zky